How can the ideal gas law be derived from the following assumptions/observations/postulates, and these only ?
- I'm able to measure pressure $P$ and volume $V$ for gases.
- I notices that if two systems of gases come into thermodynamical equilibrium, that the quantities $PV$ for gas 1 and gas 2 coincide.
- I assume there is an energy function $U$.
- The first and the second law of thermodynamics hold by axiom, together with all the consequences, which can abstractly derived from the two.
- For a free expansion of my gas $(\Delta U=\Delta Q=\Delta W=P\Delta V=0)$ I find that the expression $PV$ for my gas is the same before and after the experiment.
If I'd postulate $PV=nRT$, then I'd easily see that $U(T,V)$ is only a function of $T$ and I can derive everything else, pretty much even without the second law.
Here I know of the second law, which gives me a bunch of expressions and I wonder if that suffices to identify the temperature.
If it's not possible, are there other experiments I can do in my position which would do the job?
Here is a weaker version of the question:
Can the ideal gas law be derived if I additionally know that $U$ is really only a function of $T$, or even that $U(T,V)=C_V T\ $?
And what role does the zeroth law of thermodynamics play here?
The essence of the question is not necessarily gas physics. It's rather about how temperature is defined by the laws of Thermodynamics, if there is merely the assumption that the system can be described by internal energy $U$ and is experimentally accessible only by its parameters like $P,V,M,H,\dots$. And there is no connection to a theory lying above.
Best Answer
According to the second law, thermal equilibrium between two systems means that they both have the same temperature $T$. The fact 2 that $PV$ coincide whenever two gases are at thermal equilibrium (and, I assume, for the same $n$) means that $PV$ is only a function of $T$. In othor words, there is a function $g()$ such that $$\begin{align}PV&=g(T)& P=\frac{g(T)}{V}\end{align}$$
The goal is now to show that $g()$ is linear, i.e. that $g(T)=Tg'(T)$. In order to show that, one can use a Maxwell relation, more specifically, the one which is linked to Helmholtz free energy $A=U-TS$ which can be defined because of condition 3. We have then (including condition 4 for the computation of the derivatives) $$\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V = -\frac{\partial^2 A }{\partial T \partial V}$$
From the first equation, we have $$\left(\frac{\partial P}{\partial T}\right)_V = \frac{g'(T)}{V}.$$
Except if $PV=g(T)$ is constant over a range of temperatures (which can be checked experimentally), the condition 5 implies that $U(T,V)=U(T)$. If one make as mall isotherm transformation of our gas, one has $$\begin{align} 0=dU&=\delta Q - PdV& \delta Q=PdV\end{align}.$$ The second principle tells us then that $$dS=\frac{\delta Q}{T}=\frac{P dV}T=\frac{g(T)}{TV}dV.$$ By definition, in this case $dS=\left(\frac{\partial S}{\partial V}\right)_TdV$, so we get this partial derivative from the last equation.. Equating the two partial derivatives according to the Maxwell equation then gives us $g(T)=T g'(T)$, which implies $g'(T)=nR$, where $nR$ is an "arbitrary" constant. Hence, $$PV=nRT.$$